Is "Suppose that 0' in V satisfies 0'+x=x for each x in V" the same as "Suppose there exists 0' in V such that 0'+x=x for each x in V".

Question

Do they have the same truth set?

When proving that 0 in V is unique

I choose a similar approach as the proof in Corollary at page 554

it starts with

Suppose that 0' in V satisfies 0'+x=x for each x in V

then

0'+x=0+x

which ends with 0'=0 by cancelation law

My question is whether or not

Suppose that 0' in V satisfies 0'+x=x for each x in V

is the same as

Suppose that there exists 0' in V such that 0'+x=x for each x in V

?

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I think I seem to begin realize the meaning of the proof.

There does not exist $0'$ such that ($0'$+$x$=$x$ and $0'$=/=$0$)

is equivalent to

For all $0'$, (if $0'$+$x$=$x$, then $0'$=$0$)

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Answer 1

It is not entirely the same. In the first sentence, you are already taking for granted that $0'\in V$. Whereas in the second one, you are positing the existence of an element $0'\in V$.

However, the referenced proof is correct. The proof shows that if any element of $V$ satisfies the role of the additive identity, then that element is the identity itself. But the sentences are not the same as one already assumes such an element exists and the other posits the existence of such an element.